Reversible Reactions and Dynamic Equilibrium — AQA A-level Chemistry

SPEC 3.1.6.1 • CHEMICAL EQUILIBRIA

Dynamic Equilibrium & Le Chatelier's Principle

Understanding reversible reactions, equilibrium characteristics, and predicting shifts in equilibrium position

Reversible Reactions

Definition

A reversible reaction is a reaction that can proceed in both directions - reactants can form products (forward reaction) AND products can re-form reactants (backward reaction).

Most chemical reactions you have studied so far are presented as if they go to completion - all reactants convert to products. However, many reactions are actually reversible. The reactants can form products, but those products can also react to regenerate the original reactants.

We use a special symbol to show reversible reactions: the equilibrium arrow ⇌ (two half-arrows pointing in opposite directions). This replaces the single arrow → used for reactions that go to completion.

N₂O₄(g) ⇌ 2NO₂(g)

In this example, dinitrogen tetroxide (N₂O₄, colourless) can decompose to form nitrogen dioxide (NO₂, brown). But the NO₂ molecules can also recombine to form N₂O₄. Both processes happen simultaneously.

Key Concept

The forward reaction is left to right (N₂O₄ → 2NO₂). The backward reaction is right to left (2NO₂ → N₂O₄). In a reversible reaction, BOTH directions occur at the same time.

What is Dynamic Equilibrium?

Definition

Dynamic equilibrium is established when the rate of the forward reaction equals the rate of the backward reaction, in a closed system, resulting in constant concentrations of reactants and products.

The word "dynamic" is crucial here. At equilibrium, the reaction has NOT stopped. Both the forward and backward reactions are still occurring continuously - but they happen at exactly the same rate, so there is no overall change in concentrations. The reactants and products are constantly moving and reacting.

Equilibrium can be reached from either direction - starting with only reactants, only products, or a mixture of both. The same equilibrium position is reached regardless of which direction you approach from.

CharacteristicDescription

Closed systemNo substances can enter or leave the reaction vessel

Rate of forward = Rate of backwardBoth reactions occur at equal rates

Constant concentrationsAmounts of reactants and products do not change

Constant macroscopic propertiesColour, pressure, density remain unchanged

Common Mistake

Equilibrium does NOT mean the concentrations of reactants and products are equal. It means they are constant. A reaction might reach equilibrium with 90% products and 10% reactants, or vice versa.

Closed vs Open Systems

Key Requirement

Dynamic equilibrium can only be achieved in a closed system - where no substances can enter or leave the reaction mixture.

Closed System

No reactants or products can escape.

Equilibrium can be established.

Example: A sealed container with gases.

Open System

Matter and energy can be lost to surroundings.

Equilibrium cannot be established.

Reaction may go to completion.

CaCO₃(s) ⇌ CaO(s) + CO₂(g)

Consider the thermal decomposition of calcium carbonate:

In a closed system: The CO₂ cannot escape, so it can react with CaO to reform CaCO₃. Equilibrium is established.

In an open system: The CO₂ escapes to the atmosphere. The backward reaction cannot occur, so the reaction goes to completion - all CaCO₃ decomposes.

Note

When a reaction takes place entirely in solution (aqueous), equilibrium can be reached in open flasks because negligible material is lost through evaporation. If gases are involved, a closed system is essential.

How Equilibrium is Established

When a reversible reaction begins in a closed system, the forward reaction starts fast (lots of reactants) while the backward reaction is slow (few products). Over time, these rates change until they become equal.

Understanding the Graph

Before equilibrium: Forward rate decreases (reactants used up), backward rate increases (products accumulate). At equilibrium: Both rates are equal and constant. The reaction continues but with no net change.

Concentration Changes Over Time

We can also visualise equilibrium by plotting how concentrations change. Starting with only reactants, watch how the system evolves:

Exam Tip

The flat portions of the curves show equilibrium has been reached. Notice that concentrations are CONSTANT but NOT EQUAL - the reactant line stays higher than the product line in this example. The position depends on the value of Kc.

Le Chatelier's Principle

If a system at equilibrium is subjected to a change, the system will adjust to minimise the effect of that change and establish a new equilibrium position.

Henri Louis Le Chatelier, a French chemist, proposed this principle in 1884. It provides a powerful tool for predicting how equilibrium systems respond to changes in conditions. The key idea is that the system "opposes" the change - it shifts in the direction that counteracts what you did.

Think of it like a stubborn system that doesn't want to be disturbed. If you add more of something, it will try to use it up. If you remove something, it will try to make more. If you heat it up, it will try to cool down.

The Three Main Factors

Equilibrium position can be affected by changes in:

Concentration - adding or removing reactants/products

Pressure - for gaseous equilibria with different moles of gas on each side

Temperature - heating or cooling the system

Important Note

When equilibrium shifts, we say it "favours" the forward or backward reaction. This means MORE of that reaction occurs until a new equilibrium is established - the reactions don't stop.

Position of Equilibrium

Definition

The position of equilibrium refers to the relative amounts of reactants and products present at equilibrium. It indicates whether the equilibrium favours the formation of products (right) or the reformation of reactants (left).

Position Lies to the Right

The concentration of products is higher than that of reactants.

Equilibrium favours the forward reaction.

More product molecules present at equilibrium.

Position Lies to the Left

The concentration of reactants is higher than that of products.

Equilibrium favours the backward reaction.

More reactant molecules present at equilibrium.

Shifting Equilibrium

Shifting "to the right" means more products form (forward reaction favoured).

Shifting "to the left" means more reactants form (backward reaction favoured).

Industrial Importance

Le Chatelier's principle allows us to alter reaction conditions to maximise the yield of a desired product. This is crucial for industrial processes.

Effect of Concentration Changes

Consider the equilibrium: A + B ⇌ C + D

When you change the concentration of any species in an equilibrium mixture, the system shifts to oppose that change. The principle is straightforward: add something, and the system uses it up; remove something, and the system makes more.

Increase Concentration

Add more reactant (A or B):

Equilibrium shifts RIGHT (forward) to use up the added reactant and make more products.

Add more product (C or D):

Equilibrium shifts LEFT (backward) to use up the added product and make more reactants.

Decrease Concentration

Remove reactant (A or B):

Equilibrium shifts LEFT (backward) to replace the removed reactant.

Remove product (C or D):

Equilibrium shifts RIGHT (forward) to replace the removed product.

Industrial Application

In industry, products are often continuously removed from equilibrium mixtures. This shifts the equilibrium to make more product, improving yield. This is used in the Haber process and Contact process.

Adding Water to Aqueous Equilibria

Adding water to an aqueous equilibrium dilutes all species equally. Since both reactants and products are diluted by the same factor, there is no change in the ratio of concentrations and no shift in equilibrium position.

Worked Example01

Consider the esterification equilibrium:

CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)

(ethanoic acid + ethanol ⇌ ethyl ethanoate + water)

Predict and explain the effect on the equilibrium position when:

(a) More ethanoic acid is added

(b) The ester is removed by distillation

(c) A drying agent removes water from the mixture

(a)

More ethanoic acid is added: The system has excess reactant. To oppose this, equilibrium shifts RIGHT (forward) to use up the added acid. More ester and water are produced.

(b)

Ester is removed: The system has less product. To oppose this, equilibrium shifts RIGHT (forward) to replace the removed ester. More ester is produced (improving yield).

(c)

Water is removed: Water is a product. Removing it causes equilibrium to shift RIGHT (forward) to replace the removed water. More ester is produced. This is why drying agents improve ester yield.

Effect of Pressure Changes

Pressure changes only affect equilibria involving gases, and only when there are different numbers of moles of gas on each side of the equation.

The Rule

When pressure increases, the equilibrium shifts towards the side with fewer moles of gas (to reduce pressure). When pressure decreases, equilibrium shifts towards the side with more moles of gas.

Why does this happen? Pressure is directly related to the number of gas particles. If the system can reduce the number of particles, it reduces the pressure. This is how it "opposes" the pressure change.

N₂O₄(g) ⇌ 2NO₂(g) (colourless) (brown)

SideMoles of GasColour

Left (N₂O₄)1 moleColourless

Right (NO₂)2 molesBrown

For This Example

Increase pressure: Shifts LEFT (towards 1 mole - fewer particles). Gas becomes paler.

Decrease pressure: Shifts RIGHT (towards 2 moles - more particles). Gas becomes browner.

When Pressure Has No Effect

If the number of moles of gas is the same on both sides, changing pressure has NO EFFECT on equilibrium position.

H₂(g) + I₂(g) ⇌ 2HI(g)

SideMoles of Gas

Left (reactants)1 + 1 = 2 moles

Right (products)2 moles

Since both sides have 2 moles of gas, the system cannot reduce pressure by shifting either way. Changing pressure will NOT affect the equilibrium position.

Common Mistake

Only count gaseous species when considering pressure effects. Solids and liquids have negligible volume compared to gases, so they don't contribute to pressure. For example, in CaCO₃(s) ⇌ CaO(s) + CO₂(g), only count the 1 mole of CO₂.

Exam Tip

Always write out the mole count for each side when answering pressure questions. This shows the examiner your reasoning and helps you avoid mistakes.

Worked Example02

For the Haber process equilibrium:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

(a) How many moles of gas are on each side?

(b) Predict the effect of increasing pressure on equilibrium position.

(c) Explain why high pressure is used industrially (150-300 atm).

(a)

Left side: 1 + 3 = 4 moles of gas

Right side: 2 moles of gas

(b)

Increasing pressure causes equilibrium to shift RIGHT (forward), producing more ammonia.

(c)

The right side has fewer moles of gas (2) than the left side (4). High pressure shifts equilibrium towards fewer moles, giving a higher yield of ammonia. 150-300 atm is a compromise - higher pressures would give even better yield but require expensive high-pressure equipment.

Effect of Temperature Changes

Temperature is unique because it is the only factor that changes the value of Kc (the equilibrium constant - covered in Lesson 6.02). To predict temperature effects, you must know whether the forward reaction is exothermic or endothermic.

The Rule

Increasing temperature favours the endothermic direction (the system absorbs heat to cool down).

Decreasing temperature favours the exothermic direction (the system releases heat to warm up).

In every reversible reaction, one direction is exothermic and the opposite direction is endothermic. The diagram below shows this for the Haber process:

Increase Temperature

System absorbs heat by favouring the endothermic direction (backward).

Equilibrium shifts LEFT.

Less NH₃, more N₂ and H₂.

Decrease Temperature

System releases heat by favouring the exothermic direction (forward).

Equilibrium shifts RIGHT.

More NH₃ produced.

Exam Tip

In exam questions, the ΔH of the forward reaction will be given, so you can identify which direction is endothermic (+ve ΔH) and which is exothermic (−ve ΔH).

Temperature Changes the Value of Kc

Unlike concentration and pressure changes, temperature changes actually alter the equilibrium constant Kc. This is a crucial distinction for exams.

Forward ReactionTemperature IncreaseEffect on Kc

Exothermic (ΔH negative)Shifts backward (less products)Kc decreases

Endothermic (ΔH positive)Shifts forward (more products)Kc increases

Memory Aid

For temperature: "Heat helps endothermic" - increasing temperature always favours the endothermic direction.

Important Distinction

Concentration and pressure changes: Kc stays the same. The system adjusts concentrations to maintain the same Kc value.

Temperature changes: Kc itself changes to a new value.

Worked Example03

Consider the equilibrium:

N₂O₄(g) ⇌ 2NO₂(g) ΔH = +57 kJ mol⁻¹

N₂O₄ is colourless; NO₂ is brown.

(a) Predict the colour change when the mixture is heated.

(b) Explain your answer using Le Chatelier's principle.

(c) What happens to the value of Kc?

(a)

The mixture becomes darker brown.

(b)

The forward reaction is endothermic (ΔH = +57 kJ mol⁻¹). When temperature increases, the system opposes this by absorbing heat. This favours the endothermic (forward) direction. More NO₂ is produced, making the mixture darker brown.

(c)

Kc increases. The forward reaction is endothermic, so increasing temperature increases Kc (equilibrium shifts to give more products relative to reactants).

Effect of Catalysts

Key Fact

A catalyst has NO EFFECT on the position of equilibrium. It does NOT change the equilibrium concentrations or the value of Kc.

A catalyst is a substance that increases the rate of a chemical reaction. Crucially, a catalyst increases the rate of both the forward AND reverse reactions equally. Because both rates increase by the same factor, the ratio of rates is unchanged, and the equilibrium position stays the same.

The catalyst provides an alternative reaction pathway with lower activation energy. Since both forward and backward reactions use this same lower-energy pathway, both are sped up equally.

What Catalysts Do

A catalyst only causes a reaction to reach equilibrium faster (t₁ instead of t₂). The final equilibrium position is exactly the same - the same amounts of products and reactants are present once equilibrium is reached.

Industrial Importance

Catalysts are vital in industry because reaching equilibrium faster means faster production. Combined with removing products (which does shift equilibrium), catalysts dramatically improve efficiency.

Summary: Factors Affecting Equilibrium

FactorChangeEffect on PositionEffect on Kc

ConcentrationAdd reactant / Remove productShifts RIGHT (forward)No change

Remove reactant / Add productShifts LEFT (backward)No change

Pressure (gases only)Increase pressureShifts to fewer moles of gasNo change

Decrease pressureShifts to more moles of gasNo change

TemperatureIncrease temperatureShifts in endothermic directionChanges

Decrease temperatureShifts in exothermic directionChanges

CatalystAdd catalystNO EFFECTNo change

Exam Reminder

Temperature is the ONLY factor that changes Kc. For concentration and pressure changes, Kc stays constant - the system just adjusts to maintain it.

Practice

Question 1: Ethanol is manufactured by the hydration of ethene:

C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g) ΔH = −46 kJ mol⁻¹

(a) Explain the effect of increasing pressure on the yield of ethanol.

(b) Explain why a moderate temperature of 300°C is used rather than a lower temperature.

(c) A phosphoric acid catalyst is used. What effect does this have on the equilibrium yield?

(a)

Left side has 2 moles of gas (1 + 1); right side has 1 mole of gas. Increasing pressure shifts equilibrium RIGHT (towards fewer moles), producing more ethanol (higher yield). Industrially, ~65 atm is used.

(b)

Forward reaction is exothermic (ΔH negative). Lower temperature would favour the forward direction, giving higher yield. However, low temperature means slower rate - equilibrium takes too long to reach. 300°C is a compromise giving acceptable yield (~5%) at a practical rate.

(c)

No effect on yield. A catalyst does not change equilibrium position. It only makes equilibrium reached faster, improving production rate without changing the final yield.

Practice

Question 2: Consider the equilibrium:

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) ΔH = +124 kJ mol⁻¹

Predict and explain the effect on the equilibrium position of:

(a) Increasing the concentration of Cl₂

(b) Decreasing the pressure

(c) Increasing the temperature

(a)

Adding Cl₂ (a product) causes equilibrium to shift LEFT (backward) to use up the added Cl₂. More PCl₅ is formed; less PCl₃ and Cl₂ at new equilibrium.

(b)

Left: 1 mole of gas. Right: 1 + 1 = 2 moles of gas.

Decreasing pressure shifts equilibrium to the side with MORE moles of gas. Equilibrium shifts RIGHT (forward). More PCl₃ and Cl₂ produced.

(c)

Forward reaction is endothermic (ΔH positive). Increasing temperature favours the endothermic direction. Equilibrium shifts RIGHT (forward). More PCl₃ and Cl₂ produced. Kc increases.

Practice

Question 3: Sulfur trioxide is produced in the Contact process:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −197 kJ mol⁻¹

The industrial conditions are 450°C, 1-2 atm pressure, and V₂O₅ catalyst.

(a) Why isn't a higher pressure used, given it would increase yield?

(b) Explain why the temperature is a compromise.

(c) What effect does the catalyst have on Kc?

(a)

Left: 3 moles. Right: 2 moles. Higher pressure would shift equilibrium right, but the difference in moles is small. The cost of high-pressure equipment outweighs the small gain in yield. 1-2 atm is economically optimal.

(b)

Forward reaction is exothermic. Lower temperature favours forward direction (higher yield), but rate would be too slow. 450°C is a compromise: gives acceptable yield (about 97%) while maintaining a practical rate with the catalyst.

(c)

No effect on Kc. Catalysts do not change equilibrium constant. They only increase the rate at which equilibrium is reached.

Summary

Key Definitions

TermDefinition

Dynamic equilibriumForward and backward rates are equal; concentrations constant; closed system

Le Chatelier's principleSystem opposes changes by shifting to minimise the effect

Closed systemNo substances can enter or leave

Key Exam Points

At equilibrium: rates equal, concentrations constant (NOT equal)

Concentration/pressure changes: Kc unchanged, position shifts

Temperature changes: Kc changes, position shifts

Catalysts: NO effect on position or Kc, only rate

Always count moles of GAS only for pressure effects

Always state direction (endothermic/exothermic) for temperature effects